Chemistry, 22.07.2019 18:00, ttwright24

Abomb calorimeter has a heat capacity of 2.47 kj/k. when a 0.105-g sample of ethylene (c2h4) was burned in this calorimeter, the temperature increased by 2.14 k. calculate the energy of combustion for one mole of ethylene. a. –0.259 kj/mol b. –50.3 kj/mol c. –5.29 kj/mol d. –1.41 × 103 kj/mol e. –660 kj/mol

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answered: kisses1131953

b.-50.3 kj/mol. a heat capacity of 2.47 kJ/K tells you that you need to dd 2.47 kJ of heat to the calorimeter in order to increase its temperature by 1 K. The expression is :q=C*Î”T, the heat absorbed by the calorimeter is 2.47 kJ/K * 2.14K = 5.29 kJ. the heat released by the combustion reaction will be equal to the heat absorbed by the calorimeter. the combustion of 0.105 g of ethylene, C2H4, will give off 5.29 kJ of heat. 0.105g/5.29kJ = 1g/x 5.29kJ/.105g=x/1g x=5.29kJ/.105= 50.38kJ Since its combustion its -5.29kJ

answered: kim95207

1,41 x 10³ kJ/mol

Further explanation

Given:

A bomb calorimeter has a heat capacity of C = 2.47 kJ/K. A 0.105-g sample of ethylene (C₂H₄) was burned in this calorimeter. The temperature increased by ΔT = +2.14 K.

Question:

Calculate the enthalpy change per mole of ethylene combusted.

The Process:

Step-1: find out the amount of heat energy (Q) transferred

We will solve the problem with calorimetry calculations. We all see that the bomb calorimeter has a thermal (or thermal) capacity. During the sample was burned, there was an increase in temperature.

Q = heat capacity x temperature difference

$\boxed{ \ Q = C \ \Delta T \ }$

$\boxed{ \ Q = 2.47 \ \frac{kJ}{K} \times 2.14 \ K \ }$

Hence the amount of heat energy transferred in the reaction of the bomb calorimeter is 5.2858 kJ.

Step-2: calculate the number of moles of ethylene combusted

Molar mass of ethylene (C₂H₄) = (2 x 12.011) + (4 x 1.008) = 28.054.

Let us convert the number of grams of a sample of ethylene into moles.

$\boxed{ \ moles (n) = \frac{mass}{molar \ mass} \ }$

$\boxed{ \ moles (n) = \frac{0.105}{28.054} \ }$

We get $\boxed{ \ 3.74 \times 10^{-3} \ moles \ }$ of ethylene.

Step-3: calculate the enthalpy change per mole of ethylene combusted

$\boxed{ \ \Delta H = - \frac{heat \ transferred}{moles \ of \ substance}\ \ } \rightarrow \boxed{ \ \Delta H = - \frac{Q}{n} \ }$

$\boxed{ \ \Delta H = - \frac{5.2858 \ kJ}{3.74 \times 10^{-3} \ moles} \ }$

Thus the enthalpy change per mole of ethylene combusted is $\boxed{ \ \Delta H = 1413.32 \ kJ/mol \ } \rightarrow \boxed{ \ \Delta H \approx 1,41 \times 10^3 \ kJ/mol \ }$

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Notes:

The enthalpy change for a reaction can be measured experimentally by using a calorimeter. In a simple calorimeter all the heat evolved in exothermic reaction when is used to raise the temperature of a known mass of water (or heat capacity of a bomb calorimeter). For endothermic reactions the heat transferred from the water to the reaction can be calculated by measuring the lowering of temperature of a known mass of water (or heat capacity of a bomb calorimeter). Learn moreCalculate the work in kilojoules done during the synthesis of NH₃ link How much energy is required to vaporize 98.6 g of ethanol (C₂H₅OH) at its boiling point linkAt STP, what is the volume of nitrogen gas? link
Abomb calorimeter has a heat capacity of 2.47 kj/k. when a 0.105-g sample of ethylene (c2h4) was bur